Question

A solution is prepared by dissolving 50.4g sucrose (C₁₂H₂₂O₁₁) in 0.332kg of water. The final volume of thesolution is 355 mL. Calculate the concentration of the solution in each unit?

Molarity

Molality

Percent by mass

Mole fraction

mole percent.

Answer 1

ANSWER:

Given,

weight of sucrose = 50.4g

mol. weight of sucrose = 342.2965 g/mol

weight of water = 0.332kg = 332g

volume of solution = 355 mL

Number of moles of sucrose = {weight of sucrose}/{mol. weight of sucrose}

= {50.4g}/{342.296 g/mol}

= 0.147 mol

Number of moles of water = {weight of water}/{mol. weight of water}

= {332g}/{18g/mol}

= 18.44 mol

Now,

Molarity = {no. of mole of solute x 1000}/{volume of solution in mL}

= {0.147 mol x 1000}/{335 mL}

= 0.439 mol/L

= 0.439 M

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Molality = {no. of mole of solute x 1000}/{weight of solvent in gram}

= {0.147 mol x 1000}/{332 g}

= 0.443 mol/kg

= 0.443 m

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Percent by mass of sucrose = {mass of sucrose x 100}/{mass of sucrose + mass of water}

= {50.4 g x 100}/{50.4 g + 332 g}

= 13.18 %

Percent by mass of water = {mass of water x 100}/{mass of sucrose + mass of water}

= {332 g x 100}/{50.4 g + 332 g}

= 86.82 %

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Mole fraction of sucrose = {no. of mole of sucrose}/{no. of mole of sucrose + no. of mole water}

= {0.147 mol}/{0.147 mol + 18.44 mol}

= 0.0079

Mole fraction of water = {no. of mole of water}/{no. of mole of sucrose + no. of mole water}

= {18.44 mol}/{0.147 mol + 18.44 mol}

= 0.9921

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Mole percent of sucrose = {no. of mole of sucrose x 100}/{no. of mole of sucrose + no. of mole water}

= {0.147 mol x 100}/{0.147 mol + 18.44 mol}

= 0.79%

Mole percent of water = {no. of mole of water x 100}/{no. of mole of sucrose + no. of mole water}

= {18.44 mol x 100}/{0.147 mol + 18.44 mol}

= 99.21%